Answer Hint:When a body is taken at a depth from the surface of the Earth then the acceleration due to gravity also changes. Hence, we have to find the acceleration due to gravity at that depth. Then, we will find the weight of the body at that height and even get the decreased weight by subtracting it from the weight at the surface. Complete step by step answer:
Hence, the weight decreases by $4.6{\text{ }}N$ when we take the body to a mine of depth $5000{\text{ }}m$. Note: It must be noted that we must consider the amount of mass of the earth whose radius would be the distance between the centre of the Earth and the position of the body in which it is placed. Weight of a body is the product of the acceleration due to gravity at that place to the mass of the body. Hence, the mass of a body never changes while the weight changes from place to place.
Get the answer to your homework problem. Try Numerade free for 7 days
A body weight 500 N on the surface of the earth. How much would it weigh half way below the surface of the earth Text Solution Solution : Given that `m = 600` kg, `d = 5000 m`, `R = 6400` km `= 6.4 xx 10^(6) m`. <br> Weight of the body on the surface of the Earth `= 600 xx 9.8 = 5880 N` <br> At depth d, gravitational acceleration is <br> `g_(d) = g[1-d/R]` <br> `g_(d) = g[1-(5000)/(6.4 xx 10^(6))]` <br> `:. g_(d) = g[1-(5)/(6400)] = 9.8xx 0.999` <br> `:. g_(d) = 9.7902 m//s^(2)` <br> Weight of the body at depth <br> `d = mg_(d)` <br> `= 600 xx 9.7902` <br> `= 5874.1N` <br> `:.` Decrease in weight `= mg - mg_(d) = 5880 - 5874.1` <br> `:.` Decreases in weight `= 5.9 N` Given that m = 600 kg, d = 5000 m, At depth d, gravitation acceleration is `g_d=g[1-d/R]` `thereforeg_d=g[1-5/6400]=9.8 x 0.999` `therefore g_d=9.7902`m/s2 Weight on surface = mg =600 x 9.8 ∴Weight on surface= 5880N weight of the body at depth=mgd =600 x 9.7902 =5874N ∴Decrease in weight = mg -mgd =5880 N - 5874 N ∴Decrease in weight = 6N |